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User blog:GamesFan2000/My New Attempt At Array Notation (Part 1: Up To Three Entries)
I'm once again throwing my hat into the ring to create an array notation. This post obsoletes EPAN and will supercede it in my multi-axis array notation. I call this Hyper-Nova-Exploding Array Notation, or HNEAN. Single-Entry and Two-Entry Arrays So, let's get single entry arrays out of the way first. {1} = 1 {10^100} = 10^100 {a} = a Okay, now no one will yell at me. Let's move on to the two entry arrays. {a, b} = b layers of a-length sets of a's How does that work? Well...let's just use the simplist array that does not have 1 in it to show you. {2, 2} is solved like so: in the first layer, take 2 and create a 2-length addition set of 2's, or 2+2 = 4. Take the answer of that and go on to the second layer. Take 4 and create a 4-length multiplication set of 4's, or 4*4*4*4=256. So, {2, 2} = 256. That's what you get with 2^8. Sounds scary, am I right? To clarify, if the second entry is 1, you only do the addition part. So, let's take it up a notch, shall we? {3, 3}: 3+3+3=9 9*9*9*9*9*9*9*9*9=387420489 Now you need to create a 387420489-high power tower of 387420489's. Yes, you are looking at it right. This notation is meant to make your head explode. Even if I just used 2 in the first entry, it would be a 256-high tower of 256's. Skewe's number doesn't look so big now, does it? {4, 4}: 4+4+4+4=16 16*16*16*16*16*16*16*16*16*16*16*16*16*16*16*16≈1.84467441E19 A 1.84467441E19-high power tower of 1.84467441E19's. And whatever the heck the answer to that is has to go on to the next layer, which uses tetration. (A 1.84467441E19-high power tower of 1.84467441E19's)-long set of tetrations of a 1.84467441E19-high power tower of 1.84467441E19's. Not exactly Graham's number, but still very impressive for having only 2 entries. We go up to pentation for five layers, hexation for six, heptation for seven, and so on and so forth. Three-Entry Arrays Now the real fun begins. Consider what you have just seen a prologue to the craziness that is HNEAN. We have now reached the three entry arrays. And I'm not just going to start with an array of two's either. {3, 3, 3} And how exactly do you think we solve this? {a, b, c}= b^c-planes of layers Oh yes, now we're talking. But wait, that's not all. In {3, 3, 3}, there may be 27 planes, but how you solve them...your head will explode. Remember how {3, 3} = a 387420489-high power tower of 387420489's? That's the answer to the first plane in {3, 3, 3}. The answer to the first plane...is how many layers are in the second plane. And that 387420489-high power tower of 387420489's is used in the first layer of that second plane. Once you're done with the second plane, the third plane has a number of layers equal to the answer of the second plane, the fourth plane uses the answer to the third, and so on. Whatever the answer is to the last plane is your final answer for the array. Remember the quip about Graham's number when I was solving {4, 4}? Yeah, I'm absolutely positive that it is a speck compared to my three-entry arrays. Okay, so {2, 2, 2} might not cut it, but that's still immensely bigger than any reasonable two-entry array. But, of course, we're only at three entries. The next post I make will be on arrays with four entries and arrays with five entries. And they...they are something else. Category:Blog posts